Question: Evaluate the double integral. $ \int_{-1}^1 \int_0^{y^2} x^2 + y^2 \, dx \, dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{68}{65}$ (Choice B) B $\dfrac{52}{105}$ (Choice C) C $\dfrac{49}{125}$ (Choice D) D $\dfrac{44}{85}$
Explanation: First, we evaluate the inner integral. We can substitute in the $y^2$ at the end as if it were a numerical bound. $\begin{aligned} \int_{-1}^1 \int_0^{y^2} x^2 + y^2 \, dx \, dy &= \int_{-1}^1 \left[ \dfrac{x^3}{3} + xy^2 \right]_0^{y^2} \, dy \\ \\ &= \int_{-1}^1 \dfrac{y^6}{3} + y^4 \, dy \end{aligned}$ Second, we evaluate the outer integral. $\begin{aligned} \int_{-1}^1 \dfrac{y^6}{3} + y^4 \, dy &= \left[ \dfrac{y^7}{21} + \dfrac{y^5}{5} \right]_{-1}^1 \\ \\ &= \dfrac{1}{21} + \dfrac{1}{5} - \left( \dfrac{-1}{21} + \dfrac{-1}{5} \right) \\ \\ &= \dfrac{2}{21} + \dfrac{2}{5} \\ \\ &= \dfrac{52}{105} \end{aligned}$ The answer: $ \int_{-1}^1 \int_0^{y^2} x^2 + y^2 \, dx \, dy = \dfrac{52}{105}$